`int_0^0.6 x^2 /sqrt(9 - 25x^2) dx` Evaluate the integral

`int_0^0.6x^2/sqrt(9-25x^2)dx`

Let's first evaluate the indefinite integral by applying the integral substitution,

Let `x=3/5sin(u)`

`=>dx=3/5cos(u)du`

Plug in the above in the integral,

`intx^2/sqrt(9-25x^2)dx=int(3/5sin(u))^2/(sqrt(9-25(3/5sin(u))^2))(3/5cos(u))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9-9sin^2(u)))du`

`=int(27sin^2(u)cos(u))/(125sqrt(9(1-sin^2(u))))du`

use the identity:`1-sin^2(x)=cos^2(x)` `<br> `

`=int(27sin^2(u)cos(u))/(125*3sqrt(cos^2(u)))du`

`=int(9sin^2(u)cos(u))/(125cos(u))du`

`=9/125intsin^2(u)du`

Now use the identity:`sin^2(x)=(1-cos(2x))/2`

`=9/125int(1-cos(2u))/2du`

`=9/250int(1-cos(2u))du`

`=9/250(int1du-intcos(2u)du)`

`=9/250(u-sin(2u)/2)`

We have taken `x=3/5sin(u)`

`=>u=arcsin(5/3x)`

Substitute back u and add a constant C to the solution,

`=9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))+C`

Now let's evaluate the definite integral,

`int_0^0.6x^2/sqrt(9-25x^2)dx=[9/250(arcsin(5/3x)-1/2sin(2arcsin(5/3x)))]_0^0.6`

`=[9/250(arcsin(5/3*0.6)-1/2sin(2arcsin(5/3*0.6)))]-[9/250(arcsin(5/3*0)-1/2sin(2arcsin(5/3*0)))]`

`=[9/250(arcsin(1)-1/2sin(2arcsin(1)))]-[9/250(arcsin(0)-1/2sin(2arcsin(0)))]`

`=[9/250(pi/2-1/2sin(2*pi/2))]-[0]`

`=[9/250(pi/2-1/2sin(pi))]`

`=[9/250(pi/2-1/2*0)]`

`=(9pi)/500`

`=0.05655`

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